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Proof of the Vector Triple Product - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Vector Triple Product is considered one the most difficult concept.
25 Questions around this concept.

Solve by difficulty

The vectors $\vec{a}\; and \; \vec{b}$ are not perpendicular and $\vec{c}\; and \; \vec{d}$ are two vectors satisfying: $\vec{b}\; \times \; \vec{c}=\vec{b}\; \times \; \vec{d}$ and $\vec{a}\cdot \vec{d}=0$ . Then the vector $\vec{d}$ is equal to

$\vec{b}+\left ( \frac{\vec{b}\cdot \vec{c}}{\vec{a}\cdot \vec{b}} \right )\vec{c}$

$\vec{c}-\left ( \frac{\vec{a}\cdot \vec{c}}{\vec{a}\cdot \vec{b}} \right )\vec{b}$

$\vec{b}-\left ( \frac{\vec{b}\cdot \vec{c}}{\vec{a}\cdot \vec{b}} \right )\vec{c}$

$\vec{c}+\left ( \frac{\vec{a}\cdot \vec{c}}{\vec{a}\cdot \vec{b}} \right )\vec{b}$

View Solution

$\begin{array}{l}{\text { Let } \vec{a}=2 \hat{i}+\hat{j}-2 \hat{k} \text { and } \vec{b}=\hat{i}+\hat{j}} \\ {\text { Let } \vec{c} \text { be a vector such that }|\vec{c}-\vec{a}|=3,|(\vec{a} \times \vec{b}) \times \vec{c}|=3} \\ {\text { and the angle between } \vec{c} \text { and } \vec{a} \times \vec{b} \text { is } 30^{\circ} . \text { Then } \vec{a} . \vec{c} \text { is equal to }}\end{array}$

$2$

$5$

$\frac{1}{8}$

$\frac{25}{8}$

View Solution

Concepts Covered - 0

Vector Triple Product

For three vectors $\vec {\mathbf a},\;\vec{\mathbf b}$ and $\vec {\mathbf c}$ vector triple product is defined as $\vec {\mathbf a}\times\left ( \vec{\mathbf b}\times \vec{\mathbf c} \right )$ .
$\vec {\mathbf a}\times\left ( \vec{\mathbf b}\times \vec{\mathbf c} \right )=\left ( \vec {\mathbf a}\cdot \vec {\mathbf c} \right )\cdot \vec {\mathbf b}\;-\;\left ( \vec {\mathbf a}\cdot \vec {\mathbf b} \right )\cdot \vec {\mathbf c}$

$\\\vec p=\vec { a}\times\left ( \vec{ b}\times \vec{ c} \right )\;\text{ is a vector perpendicular to }\;\vec a\;\;\text{and }\;\;\vec{b}\times \vec {c},\;\text{ but }\;\vec{ b}\times \vec{ c} \\\text{is a vector perpendicular to the plane of }\;\vec b\;\;\text{and}\;\;\vec c.\\\text {Hence, vector } \;\vec{p} \;\text { must lie in the plane of } \vec{b} \text { and } \vec{c} \text { . }\\ \;\;\;\;\;\;\;\;\;\;\;\;Let\,\,\vec p=\vec { a}\times\left ( \vec{ b}\times \vec{ c} \right )=l\;\vec {b}+m\;\vec{c}\;\;\;\;\;\;\;\;\;\;\;\;\;[l,m\;\text{are scalars}]\;\;\;\;\;\; \ldots \text{(i)}\\\text {Taking the dot product of eq (i) with } \vec{a}, \text { we get } \\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;}\vec{p} \cdot \vec{a}=l(\vec{a} \cdot \vec{b})+m(\vec{a} \cdot \vec{c})$

$\\\mathrm{\;\;\;\;\;\;\;\;}\begin{bmatrix} \because \;\vec { a}\times\left ( \vec{ b}\times \vec{ c} \right )\;\text{is}\;\perp\;\vec a\\ \therefore \vec { a}\times\left ( \vec{ b}\times \vec{ c} \right )\cdot \vec a=0 \end{bmatrix}\\\\\text{Therefore,}\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;}\vec{p} \cdot \vec{a}=0 \\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;}l(\vec{a} \cdot \vec{b})=-m(\vec{a} \cdot \vec{c})\\\Rightarrow \;\;\;\;\;\;\;\frac{l}{\vec a \cdot\vec c}=\frac{-m}{\vec{a} \cdot \vec{b}}=\lambda\\\mathrm{{\Rightarrow \;\;\;\;\;\;\;\;\;}}l=\lambda\left ( \vec a\cdot \vec c \right )\\\mathrm{{and \;\;\;\;\;}}m=-\lambda\left ( \vec a\cdot \vec b \right )\\\text {Substituting the value of } l \text { and } m \text { in } \mathrm{Eq} \text { . (i), we get }\\\mathrm{\;\;\;}\vec{a} \times(\vec{b} \times \vec{c})=\lambda\left [ (\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a}\cdot \vec{b}) \vec{c} \right ]$

Here, the value of λ can be determined by taking specific values of $\vec{a},\;\vec{b}$ and $\vec{c}$ .

The simplest way to determine λ is by taking specific vectors $\vec{a}=\hat{i}$ , $\vec{b}=\hat{i}$ , $\vec{c}=\hat{j}$ .

We have,

$\\\mathrm{\;\;\;\;\;\;\;\;}\vec{a} \times(\vec{b} \times \vec{c})=\lambda\left [ (\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a}\cdot \vec{b}) \vec{c} \right ]\\\\\mathrm{\;\;\;\;\;\;\;\;\;}\hat{i} \times(\hat{i} \times \hat{j})=\lambda \left [ (\hat{i} \cdot \hat{j}) \hat{i}-(\hat{i} \cdot \hat{i}) \hat{j} \right ]\\\\\mathrm{\;\;\;\;\;\;\;\;\;}\hat{i} \times\hat k =\lambda \left [ (0) \hat{i}-(\1) \hat{j} \right ]\;\Rightarrow \;-\hat j=-\lambda \hat j\\\\\therefore \;\;\;\;\;\lambda=1\\\text{Hence,}\\\mathrm{\;\;\;\;\;\;\;}\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$

NOTE:

$\\1.\;\;\;\vec {\mathbf a}\times\left ( \vec{\mathbf b}\times \vec{\mathbf c} \right )=\left ( \vec {\mathbf a}\cdot \vec {\mathbf c} \right )\cdot \vec {\mathbf b}\;-\;\left ( \vec {\mathbf a}\cdot \vec {\mathbf b} \right )\cdot \vec {\mathbf c}\\\\\mathrm{\;\;\;\;\;} \left ( \vec {\mathbf a}\times \vec {\mathbf b} \right )\times\vec {\mathbf c}= \left ( \vec {\mathbf c}\cdot \vec {\mathbf a} \right )\cdot \vec {\mathbf b}\;-\;\left ( \vec {\mathbf c}\cdot \vec {\mathbf b} \right )\cdot \vec {\mathbf a} \\\\2.\;\;\text{In general}\;\;\vec {\mathbf a}\times\left ( \vec{\mathbf b}\times \vec{\mathbf c} \right )\neq\left ( \vec {\mathbf a}\times \vec {\mathbf b} \right )\times\vec {\mathbf c}\\\\\text{\;\;\;\;\;If }\;\;\vec {\mathbf a}\times\left ( \vec{\mathbf b}\times \vec{\mathbf c} \right )=\left ( \vec {\mathbf a}\times \vec {\mathbf b} \right )\times\vec {\mathbf c}\;\;\text{then the vectors }\vec a\;\;\text{and}\;\;\vec{c}\\\mathrm{\;\;\;\;\;are\;collinear}.$